answer:Given data: We have the function: begin{align}displaystyle G(x) = int_{x^2}^{e^x} ln(t^2),dt end{align} To find the derivative, we will apply the second fundamental theorem of calculus. We express the definite integral as {eq}displaystyle int_a^b g(t) dt = G(b)- G(a) {/eq} where {eq}displaystyle g(t) = ln(t^2) {/eq} and {eq}displaystyle G(t) {/eq} is its antiderivative function, with {eq}displaystyle G'(t) = g(t) {/eq}. {eq}begin{align} displaystyle G(x) &= int_{x^2}^{e^x} ln(t^2),dt[0.3cm] &=int_{x^2}^{e^x} g(t),dt [0.3cm] &= G(e^x) - G(x^2) [0.3cm] text{Take the derivative of both sides.}[0.3cm] G'(x) &= ( G(e^x) - G(x^2))'[0.3cm] text{Apply the chain rule, }(f(g(x)))' = f'(g(x))g'(x)[0.3cm] &= G'(e^x)cdot (e^x)' - G'(x^2)cdot (x^2)'[0.3cm] &= g(e^x)cdot e^x - g(x^2)cdot 2x [0.3cm] &= ln((e^x)^2)e^x - ln((x^2)^2)2x [0.3cm] &=ln(e^{2x})e^x - ln(x^4)2x [0.3cm] &=2xe^x - 4xln(x) end{align} {/eq} Therefore, the derivative is: begin{align}displaystyle mathbf{ G'(x) = 2xe^x - 4xln(x)} end{align}

answer:The inverse of the given matrix is: begin{bmatrix} -frac{20}{623} & -frac{85}{178} & frac{395}{1246} -frac{515}{1246} & -frac{75}{356} & frac{1605}{2492} -frac{565}{1246} & -frac{65}{356} & frac{1035}{2492} end{bmatrix}

answer:To prove this inequality, we'll equate it to showing that e^{frac{sqrt{5}-1}{2}} < 2, since the exponential function e^x is strictly increasing for x > 0. We can use the Taylor series expansion for e^x around x = 0: e^x = 1 + x + frac{x^2}{2!} + frac{x^3}{3!} + frac{x^4}{4!} + cdots For |x| < 1, we have an upper bound: e^x < 1 + x + frac{x^2}{2} + frac{1}{6}(x^3 + x^4 + x^5 + cdots) = 1 + x + frac{x^2}{2} + frac{1}{6}left(frac{1}{1-x} - 1 - x - frac{x^2}{2}right) Now, let x = frac{sqrt{5}-1}{2}. This value is less than 1 because sqrt{5} < 3. Substituting x into the inequality and simplifying, we get: e^{frac{sqrt{5}-1}{2}} < frac{7 + 2sqrt{5}}{6} To show that the right-hand side (RHS) is less than 2, we manipulate the fraction: frac{7 + 2sqrt{5}}{6} < 2 7 + 2sqrt{5} < 12 2sqrt{5} < 5 sqrt{5} < frac{5}{2} sqrt{5} < 2.5 Since sqrt{5} is approximately 2.236, this inequality holds true. Therefore, we've proven that ln(2) > frac{sqrt{5} - 1}{2}.

answer:The logarithm of a complex number is defined as the complex number that, when exponentiated, gives the original complex number. In other words, if z is a complex number and w is a complex number such that e^w = z, then w is the logarithm of z. To define the logarithm of a complex number, we can use the following formula: log z = ln |z| + i arg z where |z| is the modulus of z (i.e., the distance from z to the origin in the complex plane), and arg z is the argument of z (i.e., the angle between the positive real axis and the line from the origin to z). The modulus of a complex number can be found using the Pythagorean theorem: |z| = sqrt{a^2 + b^2} where a and b are the real and imaginary parts of z, respectively. The argument of a complex number can be found using the arctangent function: arg z = arctan frac{b}{a} where a and b are the real and imaginary parts of z, respectively. It is important to note that the logarithm of a complex number is not unique. This is because the argument of a complex number is not unique. For example, the argument of 1 + i is pi/4, but it is also 5pi/4, 9pi/4, and so on. As a result, the logarithm of 1 + i is also not unique. However, we can choose a unique logarithm of a complex number by restricting the argument to a specific range. For example, we can choose the argument to be in the range (-pi, pi]. This gives us the principal logarithm of a complex number. The principal logarithm of a complex number is the logarithm that has the smallest imaginary part. It is also the logarithm that is most commonly used in mathematics and physics.